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pickupchik [31]
3 years ago
9

The specific heat of aluminum is 0.90 J/gC . How much heat is given off when 25 grams of aluminum is cooled from 55 C to 25 C?

Physics
1 answer:
xxMikexx [17]3 years ago
4 0

Answer:

Q = 675 [J]

Explanation:

We can calculate the amount of heat transfer by means of the following expression that includes the mass and temperature change in a body as a function of the specific heat.

Q=m*C_{p}*(T_{initial}-T_{final})

where:

m = mass = 25 [gr]

Cp = specific heat = 0.9 [J/g*°C]

Tinitial = 55 [°C]

Tfinal = 25 [°C]

Q=25*0.9*(55-25)\\Q=675 [J]

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4.0 m/s

Explanation:

In the first part of the run, the athlete runs a distance of

d_1 = 300 m

at a speed of

v_1 = 3.0 m/s

So, the time he/she takes is

t_1 = \frac{d_1}{v_1}=\frac{300}{3.0}=100 s

In the second part of the run, the athlete covers an additional distance of

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with a speed

v_2 = 6.0 m/s

So, the time taken in this second part is

t_2 = \frac{d_2}{v_2}=\frac{300}{6.0}=50 s

So, the total distance covered is

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4 years ago
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What wave phenomena best describes the reason for the dark and light bands that
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Answer:

Interference

Explanation:

Interference is the phenomenon in which two waves superimpose to form a wave with smaller, larger or same amplitude.

There are two types of interference namely, constructive interference and destructive interference.

Constructive interference occurs when the waves are in phase and destructive interference occurs when the waves are out of phase.

In a double slit experiment, the two slits acts as sources of light and thus the waves combine to produce interference patterns. When the waves are in phase, that is the angle between them is 0°, they form a constructive interference pattern which gives rise to a light band. When the waves are out of phase, that is the angle between them is 180°, they form a destructive interference pattern which gives rise to a dark band.

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3 years ago
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C

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6 0
3 years ago
As a bicycle pump inflates a tyre, it pressure rises from 30 kPa to 40 kPa at constant temperature of 30 °C. By assuming the air
MArishka [77]

This question involves the concepts of work done, pressure, and temperature.

The work done per mol of the air is "-724.71 J/mol".

Using the given formula for work done:

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